Prove that the powerset of a finite set is finite using Coq

Question

While trying to prove some things, I encountered an innocent looking claim that I failed to prove in Coq. The claim is that for a given Finite Ensemble, the powerset is also finite. The statement is given in the Coq code below.

I looked through the Coq documentation on finite sets and facts about finite sets and powersets, but I could not find something that deconstructs the powerset into a union of subsets (such that the Union_is_finite constructor can be used). Another approach may be to show that the cardinal number of the powerset is 2^|S| but here I certainly have no idea how to approach the proof.

From Coq Require Export Sets.Ensembles.
From Coq Require Export Sets.Powerset.
From Coq Require Export Sets.Finite_sets.

Lemma powerset_finite {T} (S : Ensemble T) :
  Finite T S -> Finite (Ensemble T) (Power_set T S).
Proof.
  (* I don't know how to proceed. *)
Admitted.

I have a related result [here](https://github.com/arthuraa/extructures/blob/master/theories/fset.v), showing that the `powerset` of a finite set is also a finite set. But it relies on a different representation of finite sets, namely as ordered lists. Your result will certainly need the extensionality axiom for ensembles (`Extensionality_Ensembles`), and maybe also classical logic. — Arthur Azevedo De Amorim, May 21 '19 at 14:45
Don't know if it is relevant, but it is impossible to prove that exponentiation is total in [bounded arithmetic](https://en.wikipedia.org/wiki/Bounded_arithmetic) (I don't have a good link for that claim, but see Edward Nelson's book "Predicative Arithmetic"). This suggests that any such proof will need to use some relatively powerful principles. — John Coleman, May 21 '19 at 15:20
@ArthurAzevedoDeAmorim that is a good point. This is just part of some learning and I wouldn't mind introducing an axiom if I can then manage to proceed the proof. In general I found it interesting so far how even reasoning that other fields may consider strict or formal (such as proofs about automata) are pretty hard to translate to Coq. — Herman Bergwerf, May 21 '19 at 15:50
Learning Coq does take time. But a lot of things become easier once you get used to it. For instance, working with finite automata wouldn't be too difficult with a library of finite types and sets, along the lines of the link that I sent you. — Arthur Azevedo De Amorim, May 21 '19 at 16:29
@JohnColeman Coq's theory is quite powerful, going beyond constructive arithmetic. The issue here has more to do with the lack of extensionality for predicates, which is a pretty basic reasoning principle to add. — Arthur Azevedo De Amorim, May 21 '19 at 16:39

keep_learning · Answer 1 · 2019-05-30T15:26:03.243

I did not solve it completely because I myself struggled a lot with this proof. I merely transferred it along your line of thought. Now the crux of problem is, proving the cardinality of power set of a set of n elements is 2^n.

From Coq Require Export Sets.Ensembles.
From Coq Require Export Sets.Powerset.
From Coq Require Export Sets.Finite_sets.
From Coq Require Export Sets.Finite_sets_facts.

Fixpoint exp (n m : nat) : nat :=
  match m with
    | 0 => 1
    | S m' => n * (exp n m')
  end.

Theorem power_set_empty :
  forall (U : Type), Power_set _ (Empty_set U) = Singleton _ (Empty_set _).
Proof with auto with sets.
  intros U.
  apply Extensionality_Ensembles.
  unfold Same_set. split. 
  + unfold Included. intros x Hin.
    inversion Hin; subst. 
    apply Singleton_intro.
    symmetry. apply less_than_empty; auto.

  + unfold Included. intros x Hin.
    constructor. inversion Hin; subst.
    unfold Included; intros; assumption.
Qed.

Lemma cardinality_power_set :
  forall (U : Type) (A : Ensemble U) (n : nat),
    cardinal U A n -> cardinal _ (Power_set _ A) (exp 2 n).
Proof.
  intros U A n. revert A.
  induction n; cbn; intros He Hc.
  + inversion Hc; subst. rewrite power_set_empty.
    Search Singleton.
    rewrite <- Empty_set_zero'.
    constructor; repeat auto with sets. 
  + inversion Hc; subst; clear Hc.
Admitted.



Lemma powerset_finite {T} (S : Ensemble T) :
  Finite T S -> Finite (Ensemble T) (Power_set T S).
Proof.
  intros Hf.
  destruct (finite_cardinal _ S Hf) as [n Hc].
  eapply cardinal_finite with (n := exp 2 n).
  apply cardinality_power_set; auto.
Qed.

Prove that the powerset of a finite set is finite using Coq

1 Answers1